Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.4 - Green''s Theorem - 16.4 Exercise - Page 1102: 23

Answer

$(\dfrac{4a}{3 \pi}, \dfrac{4a}{3 \pi})$

Work Step by Step

The area of a quarter circle is: $\dfrac{1}{4} \times \ Area \ of \ circle =\dfrac{\pi a^2}{4}$ The $x$ -coordinate of the center of mass is: $\overline{x}=\dfrac{1}{2A} \oint_{C} x^2 dy \\= \dfrac{1}{\dfrac{2\pi a^2}{4}} (\dfrac{2 a^3}{3}) \\ =\dfrac{4a}{3 \pi}$ The $y$ -coordinate of the center of mass is: $\overline{y}=\dfrac{1}{2A} \oint_{C} y^2 dx \\= -\dfrac{1}{\dfrac{2\pi a^2}{4}} (-\dfrac{2 a^3}{3}) \\ =\dfrac{4a}{3 \pi}$ Thus, the center of mass is: $(\dfrac{4a}{3 \pi}, \dfrac{4a}{3 \pi})$
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